Question 1
First of all, note that linear acceleration actually always exists in circular motion, and that it always points towards the center of the circle. That is, as a particle tries to move in a straight line, it keeps on getting pulled towards the center of the circle. Try imagining a particle moving in a straight line getting pulled towards a point to the side of its path to get an intuition of this.
Another way to think about this is to compare it with orbits. Try throwing a baseball really, really fast. It falls back to the ground, right? Try throwing it even faster. It falls back to the ground, but after a longer period of time. Notice that, because the Earth is a sphere, the ground actually curves downwards. What if you threw the ball so fast, so far, that by the time gravity finally pulls it "down", the ground had already curved downwards so much that it "misses" the ground? The ball then basically is constantly falling downwards but constantly missing the ground, because the ground actually curves downwards faster than the ball can land.
In circular motion, the particle is always feeling acceleration towards the center of the circle. If this acceleration stopped, the particle would just keep on going in a straight line.
This "type" of acceleration comes up a lot in Physics. Acceleration happens everywhere, but acceleration towards a central point is a special case of acceleration. We like to give names to things, and in this case, we call it "centripetal", or "center-seeking". Makes sense, right?
To rephrase -- linear acceleration is any time the velocity of an object changes. This can happen in uniform acceleration (i mean, it's in the name...uniform acceleration is acceleration), or non uniform acceleration...any time a velocity changes, we call that change an acceleration.
Sometimes we like to study a special case of acceleration, where the acceleration always points towards a center point. We give a label to this special case of study, for convenience -- we call it "centripetal".
There isn't anything necessarily magical about this definition. It's just like us saying
"hey...it's interesting if we study acceleration always towards the center."
"hey so you what happens when there is acceleration always towards the center?"
"hey so did you hear about that acceleration always towards the center people are thinking about"
"you know if we imagine an acceleration always towards the center, you get interesting things. like circular motion."
"someone needs to think of a better name for that because it's annoying always saying 'acceleration always towards a center point' every time we talk about it."
"oh i know how about 'centripetal' acceleration idk"
"oh ok that works whatever."
Question 2
Okay so here's the thing about angular coordinates --
Every "concept" in linear coordinates has an "analog" in angular coordinates.
This shouldn't be that much of a surprise; coordinate systems are an invention of man, and the world does not care about it. There should be meaningful laws of physics no matter what coordinate systems you chose ... some just might be simpler than others.
What is "displacement"? If we think of a single dimension, it's simply the x coordinate.
What is "angular displacement"? If we think of a single dimension, it's simply the angle.
Now, what is "velocity"? It's simply the rate of change of the displacement ... the change in x over time.
What is the analogue of velocity in angular coordinates, then? It would be rate of change of the angle...the change in the angle over time.
What is "acceleration"? As we have discussed before, it's a measure of the change in velocity -- specifically, the change in velocity over time.
In the same sense, "angular acceleration" is the change in angular velocity over time.
Now, what is the "direction" of "rate of change of the angle"? There are a couple of answers to this.
- Positive, or negative -- the angle could either be getting bigger and bigger, or lower and lower.
- Clockwise or anti-clockwise -- the particle could either be moving clockwise or counter-clockwise.
So angular velocity has two "directions": clockwise, or counter-clockwise.
Now, look at a particle in circular motion and draw me a "tangent arrow" on the circle that signifies clockwise. At one point in time, the arrow might point upwards. But later on in that path, that arrow is pointing downwards. The arrow changed...but the angular velocity did not.
This is because this tangent arrow is actually the velocity of the point. It is a line in x,y coordinates. But angular velocity isn't "actually" that line. It has nothing to do with lines in space. It doesn't make any sense to think of it as a line in space. Becuase it's not a line in space.
Another way to look at this is that angular velocity and linear velocity -- clockwise/counter-clockwise and xyz -- have different "dimensions"...and putting them on the same scale is like measuring distance in kilograms, or measuring time in meters.
Obviously the arrow in xyz space does not actually represent angular velocity. Because angular velocity stayed the same, but the line changed. It is clear that the line isn't actually a good indication of the nature of angular velocity.
So what is?
Simply "counter-clockwise" or "clockwise" (or positive or negative) is the best way you can describe it, if you can establish which way you are looking at it from.
In physics however, we love arrows, so we made up some rules that we can describe counter-clockwise and clock-wise with arrows, like this:
"How do we invent a convenient way of keeping track of angular momentum...using arrows? we love arrows."
"How about we make the arrow come from the center of the circle."
"That's kind of confusing...which direction would it point at?"
"It should point perpendicular to the plane of the circle! That way nobody will actually confuse it for a real physical quantity."
"Oh yes that's right because it's just an imaginary thing we made up for convenient notation."
"Let's make it...if it's clockwise, it points down, into the page...and if it's counter-clockwise, it points up, out of the page."
"Why?"
"No real physical reason......I just like it that way."
"Ok. That's kind of cool because now we can 'add' angular velocities, the same way we can 'add' vectors. Like if we had a length 3 vector going up and we added it with a lengh 1 vector going down, we'd get a length 2 vector going up.
"Similarly if we had a counter-clockwise angular velocity of 3 and we add it with a clockwise angular velocity of 1...we get a counter-clockwise angular velocity of 2. Perfect!"
"Yes it appears that this imaginary physically meaningless notation actually makes it really convenient to add angular velocities! And it makes a lot of other things convenient too!"
Question 3
It is my hope that at by this point you know enough about the nature of angular acceleration (which is merely the change in angular velocity...and like angular velocity is either positive or negative, with no "physical" linear direction) and the "definition" of centripetal acceleration to answer your own questions.
Centripetal acceleration is a "linear", or "normal" acceleration, with x y z coordinates and everything. It has a direction. Its direction is always towards the center of the circle. Why? That's just its definition -- "centripetal" is just a name we give to acceleration towards the center of the circle.
Asking "why" is like asking why change in velocity is called acceleration. "Acceleration" is just a word we invented because we don't want to say "change in velocity" all the time.
Now I noticed you said "tangential acceleration". Be careful here. Tangential acceleration is not necessarily linear acceleration. In our case, our linear acceleration is always towards the center of the circle, and this is the case for circles of constant angular velocity. Because of this, tangential acceleration is actually zero.
Tangential acceleration starts showing up when the circular motion begins speeding up...which is something I'll leave for another tale.
Best Answer
The proper question should be "Can a varying centripetal force produce circular motion?" The answer is yes.
The (much-abused) expression $$m\frac{v^2}{r}$$ tells us what the instantaneous transverse (perpendicular) component of force must be in order for an object of mass $m$ must experience if it is travelling at instantaneous speed $v$ and is turning around an instantaneous center of curvature a distance $r$ away. It is not a force itself. It has force units, but it is the product of a mass times an acceleration. There must be some real forces which combine to result in the same net value as the product $m \times \frac{v^2}{r}$.
This is true every time the velocity of a particle changes direction.
If we apply it to circular motion, then $r$ is the radius of a circle. If we apply it to elliptical or parabolic or hyperbolic motion, then $r$ is constantly changing. If we apply it to straight line motion, then $r\rightarrow \infty$.
Back to circular motion, we can now write $${\large\Sigma}F_r=-m\frac{v^2}{r}.$$
This must be true instantaneously during every moment because the velocity direction is always changing. (The minus sign shows that the net force must be toward the center. Positive radial direction in polar coordinates is traditionally away from the center. This is important when considering a mixture of forces, some of which might, by themselves, pull the object away from the center.)
For uniform circular motion, the speed is constant, so the magnitude of the radial component of the net force must be constant. Notice that we could have two or more changing forces which still produce uniform circular motion as long as the net tangential component is zero (no speed change) and the net radial component adds up to $mv^2/r$ toward the center.
For non-uniform circular motion, we still have a constant $r$, but the speed is changing. That means that the radial component of the net force must change. How can this happen? Consider a mass attached to a string, swinging in vertical circle with non-constant speed. The mass moves fast at the bottom and slows down, but stays on the circular path at the top. The gravitational pull on the mass is constant from the Earth-bound reference frame. But it is not always a radial force. That means that the tension magnitude in the string must be changing in such a way that the net radial component always has a magnitude toward the center of $mv^2/r$.
If we define $\theta$ to be the angle measured from a vertical line passing through the center, with $\theta=0$ being at the bottom of the circle, then $$mg\cos\theta-F_\text{ten}=-m\frac{v^2}{r}$$
$$F_\text{ten}=m\left(g\cos\theta+\frac{v_{\theta}^2}{r}\right)$$ where $v_{\theta}$ is the speed at angle position $\theta$ (not the angular speed). We also know that the tension force will not affect the speed (it acts perpendicularly to the velocity), but the gravity force will change the speed like $$\frac{1}{2}mv_\text{bottom}^2=\frac{1}{2}mv_{\theta}^2 + mgr(1-\cos\theta).$$
From there it's an algebraic exercise to show that the tension force must change, and that there is a minimum required speed at the bottom for the mass to stay on the vertical circle.
Summary: for non-uniform circular motion, the radial magnitude of the net force must change, and it can.